Implicit derivation to find $\partial x/\partial v$?

Question

I saw this question:
$$\begin{cases} x^2+y^2=u \\ x\sin y+y=v\end{cases}$$

What is the $\partial x/\partial v$?

I think it should be $1/\sin(y)$ because $\partial v/\partial x=\sin y$, but the answer is $$\frac{-y}{x^2\cos y+x-y\sin y}$$
But why?

Answer 1

Your idea is correct for total derivatives, but the equivalent of

$$ \frac{\mathrm dx}{\mathrm dv}=\frac1{\frac{\mathrm dv}{\mathrm dx}} $$

is not

$$ \def\p#1#2{\frac{\partial#1}{\partial#2}} \p xv=\frac1{\p vx} $$

but the matrix equation

$$ \pmatrix{ \p xu&\p yu\\ \p xv&\p yv }= \pmatrix{ \p ux&\p vx\\ \p uy&\p vy }^{-1} $$

(which you can check by multiplying it out and applying the chain rule).

So you can indeed get the desired derivative without solving for $x$ and $y$, but you need to invert a matrix (or, equivalently, as in StackTD's answer, solve a system of linear equations).

Answer 2

Be careful: you should consider $x$ and $y$ both to be functions of $u$ and $v$. Take the partial derivative with respect to $v$ of both equations in the system to get (don't forget the chain rule!): $$\left\{ \begin{array}{rcl} \displaystyle 2x\frac{\partial x}{\partial v}+2y\frac{\partial y}{\partial v}&=&0 \\[7pt] \displaystyle \frac{\partial x}{\partial v}\sin y + x\cos y \frac{\partial y}{\partial v}+\frac{\partial y}{\partial v} &=& 1 \end{array}\right.$$ You can consider this as a system of two equations in the unknowns $\frac{\partial x}{\partial v}$ and $\frac{\partial y}{\partial v}$; solve for $\frac{\partial x}{\partial v}$.